Integration of trigonometric functions.

1. Integrals of the form $\int {\cos }^{m}x{\sin }^{n}x,dx$ are found depending on the parity of the exponents $m$ and $n$ as follows:

a) If either $m$ or $n$ is odd, then a variable substitution is used:

$t=\sin x,$ for odd $m$;

$t=\cos x,$ for odd $n,$

along with the identity ${\sin }^{2}x+{\cos }^{2}x=1;$

b) If both $m$ and $n$ are even, then reduction formulas are used:$${\sin }^{2}x=\frac{1-\cos 2x}{2},{\cos }^{2}x=\frac{1+\cos 2x}{2},\sin x\cos x=\frac{\sin 2x}{2};$$

c) If $m+n=-2k,,,,k\in \mathbb{N}$, i.e., $m+n$ is an even negative integer, then it is convenient to use substitutions

$\tan x=t$ and $\cot x=t.$

2. Integrals of the form $\int \sin mx\cos nx,dx,$ $\int \cos mx\cos nx,dx,$ $\int \sin mx\sin nx,dx$ are computed using transformations of the integrand according to the following formulas: $$\sin mx\cos nx=\frac{1}{2}(\sin (m-n)x+\sin (m+n)x),$$ $$\sin mx\sin nx=\frac{1}{2}(\cos (m-n)x-\cos (m+n)x),$$ $$\cos mx\cos nx=\frac{1}{2}(\cos (m-n)x+\cos (m+n)x).$$

3. Integrals of the form $\int {\tan }^{m}x,dx$ and $\int {\cot }^{m}x,dx$ are found using the formulas: $$t{g}^{2}x=\frac{1}{{\cos }^{2}x}-1,ct{g}^{2}x=\frac{1}{{\sin }^{2}x}-1.$$

4. Integrals of the form $R(\sin x,\cos x),dx$, where $R(u,v)$ is a rational function of two variables, are transformed into integrals of rational functions of a new variable $t$ by substituting $t=\tan \frac{x}{2}$. In this case, the following formulas are used: $$dx=\frac{2dt}{1+t^2},\sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2}.$$ If $\sin x$ and $\cos x$ under the integral contain only even powers, it is more convenient to use the substitution $t=\tan x$.

5. Integration of hyperbolic functions is carried out similarly to the integration of trigonometric functions, and the following formulas are used: $$c{h}^{2}x-s{h}^{2}x=1,shxchx=\frac{1}{2}sh2x,$$ $$c{h}^{2}x=\frac{1}{2}(ch2x+1),s{h}^{2}x=\frac{1}{2}(ch2x-1),$$ $$1-t{h}^2=\frac{1}{c{h}^{2}x},ct{h}^{2}x-1=\frac{1}{s{h}^{2}x}.$$

If $th\frac{x}{2}=t,$ then

$$shx=\frac{2t}{1-t^2};cht=\frac{1+t^2}{1-t^2};dx=\frac{2dt}{1-t^2}.$$

Examples.

Find the integrals of trigonometric functions.

1. $\int {\sin }^{3}x,dx.$

Solution.

$$\int {\sin }^{3}xdx=\int {\sin }^{2}x\cdot \sin xdx=\int (1-{\cos }^{2}x)\sin xdx=\left[\begin{array}{l}t=\cos x\\ dt=-\sin xdx\end{array}\right]=$$ $$=-\int (1-t^2)dt=-(t-\frac{t^3}{3})+C=-\cos x+\frac{{\cos }^{3}x}{3}+C.$$

Answer: $-\cos x+\frac{{\cos }^{3}x}{3}+C.$

2. $\int {\sin }^{2}x{\cos }^{2}xdx.$

Solution.

$$\int {\sin }^{2}x{\cos }^{2}xdx=\int \frac{1-\cos 2x}{2}\cdot \frac{1+\cos 2x}{2}dx=\int \frac{1-{\cos }^{2}2x}{4}dx=\frac{1}{4}\int {\sin }^{2}2xdx=$$ $$=\frac{1}{4}\int \frac{1-\cos 4x}{2}dx=\frac{1}{8}(x-\frac{1}{4}\sin 4x)+C.$$

Answer: $\frac{1}{8}(x-\frac{1}{4}\sin 4x)+C.$

3. $\int \frac{dx}{\sqrt{\cos x{\sin }^{3}x}}.$

Solution.

$$\int \frac{dx}{\sqrt{\cos x{\sin }^{3}x}}=\int \frac{1}{{\cos }^{1/2}x{\sin }^{3/2}x}dx=\int \frac{{\cos }^{3/2}x}{{\cos }^{2}x{\sin }^{3/2}x}dx=$$ $$=\int \frac{1}{t{g}^{3/2}x{\cos }^{2}x}=\left[\begin{array}{l}t=tgx\\ dt=\frac{1}{{\cos }^{2}x}dx\end{array}\right]=\int \frac{1}{t^{3/2}}dt=\int t^{-3/2}dt=$$ $$=\frac{t^{-1/2}}{-1/2}+C=-\frac{2}{\sqrt{tgx}}+C=-2\sqrt{ctgx}+C.$$

Answer: $-2\sqrt{ctgx}+C.$

4. $\int \cos \frac{x}{2}\cos \frac{x}{3}.$

Solution.

$$\int \cos \frac{x}{2}\cos \frac{x}{3}=\int \frac{1}{2}(\cos (\frac{1}{2}-\frac{1}{3})x+\cos (\frac{1}{2}+\frac{1}{3})x)dx=$$ $$=\frac{1}{2}\int (\cos \frac{1}{6}x+\cos \frac{5}{6}x)dx=\frac{6}{2}\sin \frac{1}{6}x+\frac{6}{10}\sin \frac{5}{6}x+C=$$ $$=3\sin \frac{x}{6}+0,6\sin \frac{5x}{6}+C.$$

Answer: $3\sin \frac{x}{6}+0,6\sin \frac{5x}{6}+C.$

5. $\int \sin \frac{x}{3}\cos \frac{2x}{3}dx.$

Solution.

$$\int \sin \frac{x}{3}\cos \frac{2x}{3}=\int \frac{1}{2}(\sin (\frac{1}{3}-\frac{2}{3})x+\sin (\frac{1}{3}+\frac{2}{3})x)dx=$$ $$=\frac{1}{2}\int (\sin \frac{-x}{3}+\sin x)dx=\frac{3}{2}\cos \frac{1}{3}x-\frac{1}{2}\cos x+C.$$

Answer: $\frac{3}{2}\cos \frac{x}{3}-\frac{1}{2}\cos x+C.$

6. $\int \frac{dx}{3\cos x+2}.$

Solution.

$$\int \frac{dx}{3\cos x+2}=\left[\begin{array}{l}t=tg\frac{x}{2}\\ \cos x=\frac{1-t^2}{1+t^2}\\ dx=\frac{2dt}{1+t^2}\end{array}\right]=\int \frac{2dt}{(3\frac{1-t^2}{1+t^2}+2)(1+t^2)}=$$ $$=\int \frac{2dt}{3-3t^2+2+2t^2}=2\int \frac{dt}{5-t^2}=-2\frac{1}{2\sqrt{5}}\ln \left|\frac{t-\sqrt{5}}{t+\sqrt{5}}\right|=$$ $$=\frac{1}{\sqrt{5}}\ln \left|\frac{tg\frac{x}{2}+\sqrt{5}}{tg\frac{x}{2}-\sqrt{5}}\right|.$$

Answer: $\frac{1}{\sqrt{5}}\ln \left|\frac{tg\frac{x}{2}+\sqrt{5}}{tg\frac{x}{2}-\sqrt{5}}\right|.$

7. $\int \frac{\sin x}{1+\sin x}dx.$

Solution.

$$\int \frac{\sin x}{1+\sin x}dx=\left[\begin{array}{l}t=tg\frac{x}{2}\\ \sin x=\frac{2t}{1+t^2}\\ dx=\frac{2dt}{1+t^2}\end{array}\right]=\int \frac{\frac{2t}{1+t^2}}{1+\frac{2t}{1+t^2}}\frac{2dt}{1+t^2}=$$ $$=\int \frac{4tdt}{(1+t^2)(1+2t+t^2)}=4\int \frac{tdt}{(1+t^2)(1+t{)}^2}$$ We obtained an integral of a rational function. Let's decompose the integrand into partial fractions. $$\frac{t}{(1+t^2)(1+t{)}^2}=\frac{At+B}{1+t^2}+\frac{C}{1+t}+\frac{D}{(1+t{)}^2}=$$ $$=\frac{(At+B)(1+2t+t^2)+C(1+t^2)(1+t)+D(1+t^2)}{(1+t^2)(1+t{)}^2}=$$ $$=\frac{t^3(A+C)+t^2(2A+B+C+D)+t(A+2B+C)+(B+C+D)}{(1+t^2)(1+t{)}^2}.$$ Next, by equating coefficients of like powers of $t$, we find $A$, $B$, $C$, and $D$.

$$\{\begin{array}{l}A+C=0\\ 2A+B+C+D=0\\ A+2B+C=1\\ B+C+D=0\end{array}\Rightarrow \{\begin{array}{l}A=-C\\ 2A=0\\ 2B=1\\ B+C+D=0\end{array}\Rightarrow \{\begin{array}{l}C=0\\ A=0\\ B=0,5\\ D=-0,5\end{array}$$

Thus,

$$\frac{t}{(1+t^2)(1+t{)}^2}=\frac{1}{2(1+t^2)}-\frac{1}{2(1+t{)}^2}$$ From here, we find

$$4\int \frac{tdt}{(1+t^2)(1+t{)}^2}=4(\frac{1}{2}\int \frac{1}{1+t^2}dt-\frac{1}{2}\int (1+t{)}^{-2}dt)=$$ $$=2arctgt-2\frac{(1+t{)}^{-1}}{-1}+C=2arctgt+\frac{2}{1+t}+C.$$

Making the reverse substitution, we finally obtain $$\int \frac{\sin x}{1+\sin x}dx=2arctg(tg\frac{x}{2})+\frac{2}{1+tg\frac{x}{2}}+C=2x+\frac{2}{1+tg\frac{x}{2}}+C.$$

Answer: $2x+\frac{2}{1+tg\frac{x}{2}}+C.$

8. $\int \frac{1+ctgx}{1-ctgx}dx.$

Solution.

$$\int \frac{1+ctgx}{1-ctgx}dx=\int \frac{1+\frac{\cos x}{\sin x}}{1-\frac{\cos x}{\sin x}}dx=\int \frac{\sin x+\cos x}{\sin x-\cos x}dx=$$ $$\left[\begin{array}{l}t=\sin x-\cos x\\ dt=(\cos x+\sin x)dx\end{array}\right]=\int \frac{dt}{t}dt=\ln |t|+C=\ln |\sin x-\cos x|+C.$$

Answer: $\ln |\sin x-\cos x|+C.$

9. $\int c{h}^{2}3xdx.$

Solution.

$$\int c{h}^{2}3xdx=\frac{1}{2}\int (ch6x+1)dx=\frac{1}{2}(\frac{1}{6}sh6x+x)+C=\frac{1}{12}sh6x+\frac{1}{2}x+C.$$

Answer: $\frac{1}{12}shx+\frac{1}{2}x+C.$

10. $\int \sqrt{chx+1}dx.$

Solution.

From the formula ${\cosh }^{2}x=\frac{1}{2}(\cosh 2x+1)$, we have $$\sqrt{chx+1}=\sqrt{2}ch\frac{x}{2}.$$ Therefore, $$\int \sqrt{chx+1}dx=\sqrt{2}\int ch\frac{x}{2}dx=2\sqrt{2}sh\frac{x}{2}+C.$$

Answer: $2\sqrt{2}sh\frac{x}{2}+C.$

Homework.

1. $\int \frac{{\sin }^{3}x}{{\cos }^{8}x}dx.$

2. $\int {\cos }^{7}xdx.$

3. $\int {\cos }^{2}x{\sin }^{4}xdx.$

4. $\int \frac{dx}{{\sin }^{4}x{\cos }^{2}x}.$

5. $\int \frac{\cos (x+\frac{\pi }{4})}{\sin x\cos x}dx.$

6. $\int t{g}^{3}xdx.$

7. $\int \frac{dx}{\cos \frac{x}{3}{\sin }^3\frac{x}{3}}.$

8. $\int \sin 3x\cos 5xdx.$

9. $\int \sin 10c\sin 15xdx.$

10. $\int \frac{dx}{3-2\sin x+\cos x}.$

11. $\int \frac{dx}{4{\sin }^{2}x-7{\cos }^{2}x}.$

12. $\int \frac{dx}{(\sin x+4)(\sin x-1)}.$

13. $\int s{h}^{3}2xdx.$

14. $\int \frac{dx}{s{h}^{2}xc{h}^{2}x}.$

15. $\int \frac{dx}{chx-1}.$

16. $\int ct{h}^{3}xdx.$

Tags: Integration of trigonometric functions, calculus, integral, mathematical analysis